SOLUTION: Find two numbers whose sum is 64 and whose difference is 42.

Question 246: Find two numbers whose sum is 64 and whose difference is 42.
Found 2 solutions by terrtwo, AndoyB:
Answer by terrtwo(10) (Show Source): You can put this solution on YOUR website!
The sum of two numbers can be expressed algebraicly as the sum of two variables:
x + y = 64 (equation #1)
The difference of the same two numbers is expressed in like fashion:
x - y = 42 (equation #2)
We can solve this by substitution by solving the first equation for x;
x + y = 64
--> x + y - y = 64 - y
--> x = 64 - y
We then replace the x in the second equation with (64-y):
x - y = 42
--> (64 - y) - y = 42
--> 64 -y - y = 42
--> 64 -2y = 42
We then add 2y to both sides:
--> 64 -2y + 2y = 42 + 2y
--> 64 = 42 + 2y
Then we subtract 42 from both sides:
--> 64 - 42 = 42 + 2y -42
--> 22 = 2y
Divide both sides by 2:
--> 22/2 = 2y/2
--> 11 = y
Then we go back and replace y in the first equation with 11:
x + y = 64
--> x + 11 = 64
Subtract 11 from both sides of the equation:
x + 11 -11 = 64 -11
--> x = 53
ANSWER: the two numbers are 53 and 11

Answer by AndoyB(1) (Show Source): You can put this solution on YOUR website!
53 and 11
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